A truck with growing mass

Ok, I think I might have missed a key point in this class from a few weeks ago but anyways.

There is a truck of mass M, who has a constant force F applied to it. The truck is also being filled with coal at a constant rate b kg/s. I'm supposed to find an expression for the velocity of the truck.

First I tried:

F = ma

a = F/m

Then try to integrate both sides w.r.t time to get velocity, but since the mass depends on time as well, I get an expression involving ln(m).

So then I tried it this way:

F = dp/dt

Integrate both sides,

Ft = p = mv = (M+bt)v

v = Ft/(M+bt) which is the desired result.

I think I remember something from class about the Force more accurately being the change in momentum, and not necessarily the mass-acceleration product. Was this question just to emphasis that point? I think I'll have to go read the text

Since [tex]p=mv[/tex] you have [tex]\vec{F}_{net}=\frac{d}{dt}(m\vec{v})[/tex]

If you apply the product rule you obtain [tex]\vec{F}_{net}=\vec{v}\frac{dm}{dt} \ + \ m\frac{d\vec{v}}{dt}[/tex]

If the mass is constant (ie dm/dt=0) then you are left with [tex]\vec{F}_{net}=m\frac{d\vec{v}}{dt}[/tex]. Of course [tex]\frac{d\vec{v}}{dt}[/tex] is just acceleration so you are left with [tex]\vec{F}_{net}=m\vec{a}[/tex]

Edit: Also, is the answer you obtained given as the correct answer in the book?

If you consider [tex]\vec{F}_{net}=\vec{v}\frac{dm}{dt} \ + \ m\frac{d\vec{v}}{dt}[/tex] it looks like you would get a different answer.